Eureka Airlines operates 5 daylong flight sequences serving City A, Ci

Question

Eureka Airlines operates 5 daylong flight sequences serving City A, City B, City C, City D, and City E. The following list shows the order of the cities served by each flight sequence.

Sequence 1: A-B-C-A (3 flights)
 
 Sequence 2: A-D-B-A (3 flights)
 
 Sequence 3: A-E-D-A (3 flights)
 
 Sequence 4: A-E-B-D-A (4 flights)
 
 Sequence 5: A-C-B-D-A (4 flights)

Due to the number of available flight crews, Eureka can operate only 3 flight sequences in a given day. Exactly 10 flights are flown each day. No sequence is used for more than 2 consecutive days, and no sequence is idle for 2 consecutive days.

For City A, select the number of Eureka flights that will arrive in City A over the span of any 3-day period. For City D, select the number of Eureka flights that will arrive in City D over the span of any 3-day period. Make only two selections, one in each column.

mSKR · Accepted Answer

Sequence 1: A-B-C-A (3 flights) means flight from A to B and then from B to C and then from C to A.
Question:For City A, select the number of Eureka flights that will  arrive in City A 
In S1, flights arrive in A is only 1.

As explained in  above post , you can calculate how many flights arrive in A and D.

Hope it is clear.

Sajjad1994 · Answer

Official Explanation

RO1

Each flight sequence departs from and arrives in City A one time. Exactly 10 flights are flown each day, so there must be exactly 3 flight sequences flown each day—two 3-flight sequences and one 4-flight sequence. Therefore, 3 flights must arrive in City A each day, making 9 flights that must arrive in any 3-day span.
 
  The correct answer is 9.  
 
 RO2

As explained in the analysis of RO1, one 4-flight sequence is flown each day. Each 4-flight sequence arrives in City D one time, thus accounting for 3 arrivals in City D over a 3-day span.
 
Also explained in the analysis of RO1, two 3-flight sequences are flown each day. Sequence 1 does not arrive in City D. Sequences 2 and 3 each arrive in City D once. Since no sequence can be used for more than 2 consecutive days, and no sequence is idle for 2 consecutive days, each 3-flight sequence must be used in a pattern of two days in use and one day idle. Therefore, in any 3-day span, each 3-flight sequence will be used twice. Thus, Sequences 2 and 3 each arrive in City D twice in a 3-day span, for a total of 4 arrivals. Added to the 3 arrivals accounted for by the 4-flight sequences, the total number of Eureka flights that will arrive in City D over any 3-day span is 7.
 
  The correct answer is 7.

Apt0810 · Answer

Here if we see ,in all the sequences the ending point is point A,so if we see any sequence in the given choices we will see that on each day there will three arrivals at City A,so on a 3 day period the number of arrivals at A will be =3*3=9

And if we see the second max arrivals would be in City D,so if we select any sequence to get 10 flights in a day we will get a total of 7 flights,we can also see it by taking an example where I will take the worst case when I am taking Sequence 1 for two consecutive days(since in it D is not there)Now 
for example on Day 1 - Seq1,Seq2,Seq4
Day2- Seq1,Seq3,Seq4
Day3- Seq3,Seq5,Seq2

So we will get total number of arrivals at D as 7 and we also check the case of other examples but the answer will still remain the same

Posted from my mobile device

mSKR · Answer

Span of any 3 day period under constraints
Day1: S1 S2 S4
Day2 :S1 S3 S5
Day3: S2 S3 S4

Each sequence will have 1 flight arrive in A and S1 don't have D flights

CityA: 9 flights ( A is available in each sequence) 
City D is not in S1 ( 7 flights)

scranjith · Answer

Premise: "Exactly 10 flights are flown each day."

Question: select the number of Eureka flights that will arrive in City A over the span of any 3-day period.

Since 10 flights are flown everyday, why is the answer 3 flights a day arrives at A instead of 10 flights. Which will total up to 30 flights arriving at A in a span of 3 days and 24 flights in D.

Had the question be to determine number of unique flight sequence that will arrive at A, i would have chosen the current answer.

What am in missing in this question?

Posted from my mobile device

sayan640 · Answer

MartyMurray  Can you please provide your explanation for this question ?  KarishmaB

sayan640 · Answer

It's important to understand the sequence first..In this sequence Sequence 1: A-B-C-A (3 flights) , 1 flight departs from A and arrives at B . Another flight departs from B and arrives at C . And another flight departs from C and arrives at A . So there is only one occasion when 1 flight arrives at A in this particular sequence. 
Once this is clear we can apply the same rule for other cases.

MBAToronto2024 · Answer

KarishmaB  
Can you please explain the official answer? 
Cannot understand it:

City A, sequence of (4-1-2): 3 + 3 + 4 = 10 flights. In seq 1, we'd fly three times cause we have 3 flights, using same logic 3+3+4 = 10. Why 9?

KarishmaB · Answer

Sequence 1: A-B-C-A (3 flights) - S1
Sequence 2: A-D-B-A (3 flights) - S2
Sequence 3: A-E-D-A (3 flights) - S3
Sequence 4: A-E-B-D-A (4 flights) - S4
Sequence 5: A-C-B-D-A (4 flights) - S5

Each day - 3 sequences

Exactly 10 flights each day. 10 can only be obtained in one way here: 4 + 3 + 3 flights. 
It means everyday one of S4 and S5 occur and every day 2 of S1/S2/S3 occur. 
We cannot have say S1, S2 and S3 on a day because that will make up only 9 flights.

No sequence is used for more than 2 consecutive days and no sequence is idle for 2 consecutive days.

It means if S4 is used on Day 1, S5 must be used on Day 2 because S5 cannot be idle for both day 1 and day 2. On Day 3, again S4 will be used and so on.
Also, if S1 and S2 are used on Day 1, S2 and S3 should be used on Day 2 (because S3 cannot be idle on both days). Then S1 and S3 should be used on Day 3 because S1 cannot be used for more than 2 days consecutively. Then the sequence could look something like this:

Day 1: S1, S2, S4
Day 2: S2, S3, S5
Day 3: S1, S3, S4
Day 4: S1, S2, S5
Day 5: S2, S3, S4
Day 6: S1, S3, S5
The same will be repeated on and on.

Consider Days 1, 2 and 3.
In every sequence, a flight reaches A for sure. Hence with 3 flights everyday, in 3 days there will be 9 flights to A.

Select for  City A: 9

In both S4 and S5, a flight reaches D. So in 3 days, these 3 flights will reach D.
Out of S1, S2 and S3, S2 and S3 involve flights to D. So on the 3 days (say Day 1 to Day 3), S2 and S3 will be followed twice each. Hence these 4 flights will also reach D. 
In 3 days, 7 flights will reach D. 
 
Select for  City D: 7

kayarat600 · Answer

It's not possible to fly 10 times in a day only 9 or 11

BinodBhai · Answer

This question is not difficult but time consuming..

Eureka Airlines operates 5 daylong flight sequences serving City A, Ci

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Eureka Airlines operates 5 daylong flight sequences serving City A, Ci

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